Let p(x) have coefficients a0 a1 … an in R, with an ≠ 1; hence p is not monic. Let u be a root of p(x), the algebraic element.
Build a new polynomial q(x) by multiplying ai by kn-i. This multiplies the constant by kn, and changes the lead coefficient not at all. Verify that ku is a root of q(x).
Let k = an, then divide q(x) through by an to find a monic polynomial with root anu.
We need R to be an integral domain, so that anu does not become 0.
A similar result holds if p(x) is monic and its coefficients lie in S inverse of R, where S has no zero divisors. In this case R need not be an integral domain. Let k be the product of the denominators of the coefficients of p(x), and build q(x) as above. Now ku is nonzero, and is a root of the monic polynomial q(x).